I am looking for the polynomials invariant under two linear transformations. That is, if $x\in\mathbb{R}^4$ and given two sets of linear transformations $f$ and $g$, I am looking for the invariant polynomials p(x) such that
$$
p(f(x))=p(g(x))=p(x).
$$
In my case, writing $x=(x_1,x_2,x_3,x_4)$, the transformations are $x'_i=f(x_i)=(-1)^{i}x_i$ and $g$ depends on a continuous parameter $t\in \mathbb{R}$,
$$
g_t(x)=\begin{pmatrix}
1 & 0 & 0 &0\\
2t & 1 & 0 & 0\\
3t^2 & 3t & 1 &0\\
4t^3 & 6t^2 & 4t & 1
\end{pmatrix}\begin{pmatrix}x_1\\ x_2\\ x_3\\ x_4\end{pmatrix}.
$$
Note that $g_0$ is the identity transformation, and thus $F=\{g_0,f\}$ is a group, and also, $G=\{g_t\}_{t\in\mathbb{R}}$ is a group, since $g_{t'}(g_{t'}(x))=g_{t+t'}(x)$. Thus $H=\{\{g_t\}_{t\in\mathbb{R}},f\}$ is a group, with $F$ a subgroup. We also have $f.g.f=g^{-1}$ for all $g\in G$. It turns out that $G$ is a Lie group, with $g_t=e^{t X}$ with generator
$$X=\begin{pmatrix}
0 & 0 & 0 &0\\
2 & 0 & 0 & 0\\
0 & 3 & 0 &0\\
0 & 0 & 4 & 0
\end{pmatrix}.$$
I have convinced myself that all invariant polynomials can be written as
$$p(x)=q(y)$$
with $q$ a polynomial of the "basis" $y=(x_1^2,4x_1x_3-3 x_2^2,x_1^2x_4-2x_1 x_2 x_3+x_2^3)$.
However, how do I prove this? That is, how do I prove that there are only three "irreducible" invariant polynomials and that any other invariant polynomial can be written in terms of them?
For instance, is the fact that $X^4=0$ related to the fact that there are only three irreducible invariant polynomials?
Best Answer
Using the fact that $G$ is a one-parameter Lie group (this is the relevant key-word to solve the problem) allows for the use of the literature on these problems. I follow the "Introduction to the Lie theory of one-parameter groups" by Cohen.
Using that $G$ is a simple case of a one-parameter Lie group, we see easily that the transformation induced by G on $x$ is $x(t)=e^{Xt}x$.
Since $G$ is a Lie group, we can focus on the study of infinitesimal transformations, $$ \partial_t x=Xx=Dx,\quad\quad (1) $$ where we define the differential operator $D=x X^T\partial_x=2x_1 \partial_{x_2}+3x_2 \partial_{x_3}+4x_3 \partial_{x_4}=\sum_{i=1}^4\xi_i(x) \partial_{x_i}$. The infinitesimal transformation of a generic function $u(x)$ is given by $Du$.
It is better to solve Eq. (1) as $$ \frac{dx_1}{\xi_1}=\frac{dx_2}{\xi_2}=\frac{dx_3}{\xi_3}=\frac{dx_4}{\xi_4}=dt,\quad\quad (2) $$ where in our case, the first equation should be interpreted as $dx_1=0$, and thus $x_1(t)=i_1(x(t))=c_1$ is an invariant. Indeed, Eq. (2) is very convenient, since it allows us to directly find the invariants, as $\frac{dx_2}{\xi_2}=\frac{dx_3}{\xi_3}$ and $\frac{dx_2}{\xi_2}=\frac{dx_4}{\xi_4}$ are independent of $t$. Solving these two equations gives two additional invariants $i_2(x(t))=4 x_1(t)x_3(t)-3 x_2(t)^2=c_2$ and $i_3(x(t))=x_1(t)^2x_4(t)-2x_1(t)x_2(t)x_3(t)+ x_2(t)^3=c_3$. Finally, solving for instance $\frac{dx_2}{\xi_2}=dt$ we obtain $v(x(t))=x_2(t)/2x_1(t)=c_4+t$.
Therefore, instead of expressing the transformation induced by $G$ in terms of the trajectory $x(t)$, we parametrize it in terms of the three invariants $i_1$, $i_2$ and $i_3$, as well as the simple evolution of $v(x(t))$.
A theorem then states that for an arbitrary function $u(x)$ to be invariant under $G$, it is necessary and sufficient that $Du=0$. This is equivalent to say that $\frac{d u(x(t))}{dt}=0$. But with the above parametrization, $u(x(t))=U(i_1(x(t)),i_2(x(t)),i_3(x(t)),v(x(t)))=U(i_1,i_2,i_3,c_4+t)$. Invariance of $u$ implies that any invariant function can be written as $u(x)=U(i_1,i_2,i_3)$.
We are left with the invariance under $F$. We can readily check that $f(i_1)=-i_1$ and $f(i_2)=i_2$, $f(i_3)=i_3$, which leads us to introduce the invariants under the full group $H$ as $I_1=i_1^2$, $I_2=i_2$ and $I_3=i_3$. Thus an arbitrary function $u(x)$ invariant under $H$ is of the form $u(x)=U(I_1,I_2,I_3)$